📚 CCEA Year 13 Physics Unit Test Mock Paper Walkthrough | CCEA 13年级物理单元测试模拟卷解析
This article provides a full walkthrough of a mock unit test designed for Year 13 CCEA Physics students. The paper covers core A2 topics including circular motion, simple harmonic motion, momentum, thermal physics, electric fields, nuclear decay, and experimental analysis. Each question is followed by a model answer with clear reasoning, helping you master common exam-style problems and avoid typical pitfalls.
本文为 CCEA 13 年级物理单元测试模拟卷提供详细解析。试卷涵盖圆周运动、简谐运动、动量、热物理、电场、核衰变以及实验分析等 A2 核心主题。每道题均配有标准答案和清晰的解题思路，帮助你掌握常见的考试题型并避开典型错误。
1. Overview of the Mock Paper | 模拟卷概览
The mock paper consists of seven compulsory questions totalling 70 marks, designed to be completed in 1 hour 15 minutes. Questions blend short calculations, derivations, graph interpretations, and an extended experimental design task. The topics align with the CCEA A2 Unit 1 specification, emphasising application of principles to unfamiliar contexts.
模拟卷共包含七道必答题，总分 70 分，设计用时 1 小时 15 分钟。题目融合了简短计算、公式推导、图像解读和一道拓展实验设计题。主题紧扣 CCEA A2 第一单元大纲，侧重将物理原理应用于陌生情境。
2. Question 1: Circular Motion on a Banked Track | 第1题：倾斜轨道上的圆周运动
A vehicle of mass 1200 kg travels around a banked circular track of radius 80 m. The track is banked at an angle of 25° to the horizontal. Calculate the ideal speed at which the vehicle should travel so that no lateral friction is required. Hence, determine the centripetal force acting on the vehicle at this speed.
一辆质量为 1200 kg 的汽车在半径 80 m 的倾斜圆形赛道上行驶，赛道与水平面的倾角为 25°。计算车辆在不依赖侧向摩擦力时的理想行驶速度，并求出在该速度下作用于车辆的向心力。
For a banked track with no friction, the horizontal component of the normal reaction provides the centripetal force: N sin θ = m v² / r, and vertical equilibrium gives N cos θ = m g. Dividing the equations yields tan θ = v² / (r g). Rearranging: v = √(r g tan θ). Substituting values: v = √(80 m × 9.81 m s⁻² × tan 25°) = √(80 × 9.81 × 0.4663) ≈ √366.0 = 19.1 m s⁻¹. The centripetal force F_c = m v² / r = 1200 × (19.1)² / 80 ≈ 1200 × 365 / 80 = 5475 N.
无摩擦倾斜弯道的向心力由支持力的水平分量提供：N sin θ = m v² / r，竖直方向平衡：N cos θ = m g。两式相除得 tan θ = v² / (r g)。整理得 v = √(r g tan θ)。代入数据：v = √(80 m × 9.81 m s⁻² × tan 25°) = √(80 × 9.81 × 0.4663) ≈ √366.0 = 19.1 m s⁻¹。向心力 F_c = m v² / r = 1200 × (19.1)² / 80 ≈ 1200 × 365 / 80 = 5475 N。
3. Question 2: Simple Harmonic Motion Analysis | 第2题：简谐运动分析
A 0.50 kg mass attached to a spring oscillates with an amplitude of 3.0 cm and a period of 0.80 s. Determine the spring constant k and the maximum acceleration of the mass. Sketch the acceleration–displacement graph for one complete cycle, marking key values.
一个 0.50 kg 的物体系在弹簧上作简谐振动，振幅为 3.0 cm，周期为 0.80 s。计算弹簧的劲度系数 k 及物体的最大加速度。绘制一个完整周期的加速度–位移图像，并标出关键数值。
Angular frequency ω = 2π / T = 2π / 0.80 = 7.854 rad s⁻¹. For a spring-mass system, ω = √(k / m), so k = m ω² = 0.50 × (7.854)² ≈ 30.9 N m⁻¹. Maximum acceleration a_max = ω² A, where A = 0.030 m. a_max = (7.854)² × 0.030 = 61.7 × 0.030 = 1.85 m s⁻². The a–x graph is a straight line through the origin with negative slope –ω², extending from x = –A to +A, with a ranging from +a_max to –a_max.
角频率 ω = 2π / T = 2π / 0.80 = 7.854 rad s⁻¹。对于弹簧振子，ω = √(k / m)，所以 k = m ω² = 0.50 × (7.854)² ≈ 30.9 N m⁻¹。最大加速度 a_max = ω² A，其中 A = 0.030 m。a_max = (7.854)² × 0.030 = 61.7 × 0.030 = 1.85 m s⁻²。加速度–位移图像为过原点、斜率为 –ω² 的直线，范围为 x = –A 至 +A，a 由 +a_max 到 –a_max。
4. Question 3: Momentum and Impulse | 第3题：动量与冲量
A tennis ball of mass 58 g strikes a racket horizontally at 32 m s⁻¹ and rebounds at 28 m s⁻¹ in the opposite direction. The contact time is 12 ms. Calculate the change in momentum of the ball and the average force exerted by the racket. Explain why the force calculated is an average.
一个质量为 58 g 的网球以 32 m s⁻¹ 的速度水平撞击球拍，并以 28 m s⁻¹ 的速度反向弹回，接触时间为 12 ms。计算球的动量变化量以及球拍施加的平均力。解释为什么所求的力是平均值。
Take the initial direction as positive. Initial momentum p_i = m u = 0.058 kg × 32 m s⁻¹ = 1.856 kg m s⁻¹. Final momentum p_f = 0.058 × (–28) = –1.624 kg m s⁻¹. Change in momentum Δp = p_f – p_i = –1.624 – 1.856 = –3.48 kg m s⁻¹. Magnitude of change is 3.48 kg m s⁻¹. Impulse F_avg Δt = Δp, so F_avg = Δp / Δt = 3.48 / 0.012 = 290 N. The force varies during contact, so F_avg represents the constant equivalent force that would produce the same impulse.
设定初速度方向为正。初动量 p_i = m u = 0.058 kg × 32 m s⁻¹ = 1.856 kg m s⁻¹。末动量 p_f = 0.058 × (–28) = –1.624 kg m s⁻¹。动量变化 Δp = p_f – p_i = –1.624 – 1.856 = –3.48 kg m s⁻¹，变化量大小为 3.48 kg m s⁻¹。冲量 F_avg Δt = Δp，所以 F_avg = Δp / Δt = 3.48 / 0.012 = 290 N。接触过程中力是变化的，因此 F_avg 代表能够产生相同冲量的等效恒力。
5. Question 4: Thermal Physics and Ideal Gas | 第4题：热物理与理想气体
A cylinder contains 0.25 mol of an ideal gas at a pressure of 2.0 × 10⁵ Pa and temperature 300 K. The gas is heated at constant volume until its pressure doubles. Calculate the final temperature and the work done during the process. The gas is then allowed to expand isothermally back to its original pressure. Determine the final volume if the initial volume was 3.1 × 10⁻³ m³.
一气缸装有 0.25 mol 理想气体，初始压强为 2.0 × 10⁵ Pa，温度为 300 K。在体积不变条件下对气体加热，直至压强加倍。计算末态温度及过程中气体做功。随后气体等温膨胀回到初始压强。若初始体积为 3.1 × 10⁻³ m³，求最终体积。
Constant volume: p₁/T₁ = p₂/T₂, so T₂ = T₁ × (p₂/p₁) = 300 K × 2 = 600 K. Work done W = 0 because ΔV = 0. For isothermal expansion back to p₁ = 2.0 × 10⁵ Pa, p₂_initial = 4.0 × 10⁵ Pa. Using p V = constant: p₂_initial × V₁ = p₁ × V_final, so V_final = (p₂_initial / p₁) × V₁ = 2 × 3.1 × 10⁻³ = 6.2 × 10⁻³ m³.
体积不变：p₁/T₁ = p₂/T₂，得 T₂ = T₁ × (p₂/p₁) = 300 K × 2 = 600 K。做功 W = 0，因为 ΔV = 0。随后等温膨胀回到初始压强 p₁ = 2.0 × 10⁵ Pa，此时起始压强 p₂_initial = 4.0 × 10⁵ Pa。由 pV = 常数：p₂_initial × V₁ = p₁ × V_final，因此 V_final = (p₂_initial / p₁) × V₁ = 2 × 3.1 × 10⁻³ = 6.2 × 10⁻³ m³。
6. Question 5: Uniform Electric Field | 第5题：匀强电场
Two parallel metal plates are separated by 4.0 cm and connected to a 600 V power supply. An electron enters the field midway between the plates with a horizontal speed of 3.0 × 10⁶ m s⁻¹ parallel to the plates. Calculate the electric field strength, the vertical acceleration of the electron, and the vertical displacement after travelling 5.0 cm horizontally. Ignore gravitational effects.
两块平行金属板相距 4.0 cm，连接至 600 V 电源。一个电子以平行于板的水平速度 3.0 × 10⁶ m s⁻¹ 从两极板中间进入电场。忽略重力影响，计算电场强度、电子的竖直加速度以及水平移动 5.0 cm 后的竖直位移。
Electric field E = V / d = 600 V / 0.040 m = 1.5 × 10⁴ V m⁻¹ (or N C⁻¹). Force on electron F = e E = 1.60 × 10⁻¹⁹ C × 1.5 × 10⁴ N C⁻¹ = 2.4 × 10⁻¹⁵ N. Acceleration a = F / m_e = 2.4 × 10⁻¹⁵ / 9.11 × 10⁻³¹ ≈ 2.63 × 10¹⁵ m s⁻². Time to travel horizontally t = x / v_x = 0.050 m / 3.0 × 10⁶ m s⁻¹ = 1.67 × 10⁻⁸ s. Vertical displacement y = ½ a t² = 0.5 × 2.63 × 10¹⁵ × (1.67 × 10⁻⁸)² ≈ 0.5 × 2.63 × 10¹⁵ × 2.79 × 10⁻¹⁶ ≈ 0.367 m = 3.7 cm. This exceeds half the plate separation (2.0 cm), so the electron would hit the upper plate before reaching 5.0 cm horizontally; accurate exam answers should note this and calculate the time to impact or state the limitation.
电场强度 E = V / d = 600 V / 0.040 m = 1.5 × 10⁴ V m⁻¹（或 N C⁻¹）。电子受力 F = e E = 1.60 × 10⁻¹⁹ C × 1.5 × 10⁴ N C⁻¹ = 2.4 × 10⁻¹⁵ N。加速度 a = F / m_e = 2.4 × 10⁻¹⁵ / 9.11 × 10⁻³¹ ≈ 2.63 × 10¹⁵ m s⁻²。水平运动时间 t = x / v_x = 0.050 m / 3.0 × 10⁶ m s⁻¹ = 1.67 × 10⁻⁸ s。竖直位移 y = ½ a t² = 0.5 × 2.63 × 10¹⁵ × (1.67 × 10⁻⁸)² ≈ 0.5 × 2.63 × 10¹⁵ × 2.79 × 10⁻¹⁶ ≈ 0.367 m = 3.7 cm。该值已超出板间距的一半（2.0 cm），因此电子在水平移动 5.0 cm 前会撞击上极板；准确的考试答案应指出这一点，并计算撞击时间或说明限制。
7. Question 6: Radioactive Decay and Half-life | 第6题：放射性衰变与半衰期
A sample of iodine-131 has an initial activity of 8.0 × 10⁶ Bq and a half-life of 8.0 days. Calculate the decay constant in s⁻¹ and the number of radioactive atoms initially present. Determine the activity after 24 days and sketch the activity–time graph, labelling the half-life.
一块碘-131样品的初始活度为 8.0 × 10⁶ Bq，半衰期为 8.0 天。计算衰变常量（以 s⁻¹ 为单位）以及最初存在的放射性原子数。求 24 天后的活度，并绘制活度–时间图，标出半衰期。
Half-life T₁/₂ = 8.0 days = 8.0 × 24 × 3600 s = 691200 s. Decay constant λ = ln 2 / T₁/₂ = 0.693 / 691200 ≈ 1.00 × 10⁻⁶ s⁻¹. Initial number of atoms N₀ = A₀ / λ = 8.0 × 10⁶ / 1.00 × 10⁻⁶ = 8.0 × 10¹². After 24 days, 3 half-lives have elapsed, so activity A = A₀ / 2³ = 8.0 × 10⁶ / 8 = 1.0 × 10⁶ Bq. The graph is an exponential decay curve starting at (0, 8.0×10⁶ Bq), passing through (8 days, 4.0×10⁶ Bq), (16 days, 2.0×10⁶ Bq), (24 days, 1.0×10⁶ Bq).
半衰期 T₁/₂ = 8.0 天 = 8.0 × 24 × 3600 s = 691200 s。衰变常量 λ = ln 2 / T₁/₂ = 0.693 / 691200 ≈ 1.00 × 10⁻⁶ s⁻¹。初始原子数 N₀ = A₀ / λ = 8.0 × 10⁶ / 1.00 × 10⁻⁶ = 8.0 × 10¹²。经过 24 天即 3 个半衰期，活度 A = A₀ / 2³ = 8.0 × 10⁶ / 8 = 1.0 × 10⁶ Bq。活度–时间图为指数衰减曲线，起点为 (0, 8.0×10⁶ Bq)，经过 (8 天, 4.0×10⁶ Bq)、(16 天, 2.0×10⁶ Bq)、(24 天, 1.0×10⁶ Bq)。
8. Question 7: Experimental Skills – Young Modulus | 第7题：实验技能 – 杨氏模量
Describe a procedure to determine the Young modulus of a copper wire. Include the measurements needed, the graph you would plot, and how the value is extracted. Identify the main source of uncertainty and suggest an improvement. State the typical value of Young modulus for copper.
描述测定铜丝杨氏模量的实验步骤，包括需要测量的量、绘制的图像以及如何从图像中求出数值。指出主要的不确定度来源并提出改进方法。列出铜的杨氏模量典型值。
Suspend a long, thin copper wire vertically and attach a scale to measure extension. Measure the initial length L with a metre rule and the diameter d with a micrometer screw gauge at several points to find cross-sectional area A = π d² / 4. Add known masses and record the extension ΔL for each load F = m g. Plot stress (F/A) on the y-axis against strain (ΔL/L) on the x-axis. The gradient of the linear portion gives the Young modulus E. The main uncertainty arises from measuring the small extension; use a travelling microscope or a Vernier scale to improve precision. Typical Young modulus for copper is about 1.1 × 10¹¹ Pa to 1.3 × 10¹¹ Pa. Ensure the elastic limit is not exceeded.
将一根细长的铜丝竖直悬挂，并安装刻度尺以测量伸长量。用米尺测量初始长度 L，用螺旋测微器在多点测量直径 d，计算横截面积 A = π d² / 4。逐次增加已知质量，记录每个荷载 F = m g 对应的伸长量 ΔL。以应力 (F/A) 为纵轴、应变 (ΔL/L) 为横轴作图。直线区域的斜率即为杨氏模量 E。主要不确定度来自微小伸长量的测量；可使用移测显微镜或游标卡尺来提高精度。铜的典型杨氏模量约为 1.1 × 10¹¹ Pa 到 1.3 × 10¹¹ Pa。务必确保不超过弹性极限。
9. Common Mistakes and Examiner Tips | 常见错误与考官建议
Many students lose marks by confusing angular frequency ω with angular velocity, especially in SHM where energy calculations demand ω in rad s⁻¹. Always convert units to SI before substituting, and check that radians are used for phase angles. In momentum questions, direction must be assigned consistently; a change in momentum is a vector subtraction. For thermal physics, remember that the internal energy of an ideal gas depends solely on temperature. Graph plotting requires a sharp pencil, labelled axes with units, and a line of best fit that may not pass through every point. When describing experiments, be specific about instruments and how they are used, not just their names.
许多学生因混淆角频率 ω 与角速度而失分，尤其在简谐运动能量计算中，ω 的单位必须是 rad s⁻¹。代入公式前务必统一换算为国际单位，并检查相位角是否使用弧度。在动量类问题中，必须始终规定方向，动量变化是矢量差。在热物理中牢记理想气体的内能只取决于温度。作图时需使用尖细铅笔，坐标轴标明物理量与单位，最佳拟合线不一定通过所有数据点。描述实验时，要具体说明所用仪器及其使用方法，而非仅仅提及名称。
10. Conclusion and Further Practice | 总结与进一步练习
This mock paper has reinforced key A2 concepts: circular motion on a banked track, SHM parameters, impulse-momentum, gas laws, charged particle deflection, radioactive decay, and practical determination of the Young modulus. For thorough preparation, attempt past CCEA papers under timed conditions and review mark schemes to understand the precise wording expected. Practice deriving the ideal banking equation and the exponential decay law from first principles, as these derivations often appear in longer structured questions.
本次模拟卷巩固了关键 A2 概念：倾斜弯道的圆周运动、简谐运动参数、冲量–动量定理、气体定律、带电粒子偏转、放射性衰变以及杨氏模量的实验测定。为充分备考，请在规定时间内练习历年 CCEA 真题，并研读评分标准以明确所需的标准表达。多练习从基本原理推导理想倾斜速度方程和指数衰变定律，这些推导常出现在较长的结构化题目中。
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